Here a = 9,  b = 5, c = 0
\(\therefore\)  D = b² - 4ac = 52 - 4 (9) (0) = 25
D > 0 and it is a perfect square, the roots are real and distinct

 Given y = b^x
Converting this into logarithmic form, we get

\({ log }_{ b }^{ y }\)   = x

The domain logarithmic function is the set of positive real numbers and the range is the set of real numbers.
Domain (0,\(\infty\)), and Range (0,\(\infty\))

x < -1 or x < 3
⇒ x ∈ (-∞, 3]

x < 5 and x > -3
⇒ x ∈ [-3, 5]

[a < b ⇒ ay > by for all y< 0]
Here y = -1
Given |3-x| < 7
This means -7 < 3 -x < 7
⇒-7-3 < -x < 7-3
⇒-10 < -x < 4
⇒10 > x > -4
\(\therefore\) The Solution set is \(x\in \left( -\infty ,-4 \right) \cup \left( -4,10 \right) \)

Let r be the radius of the spherical tank
 Then, Volume of the spherical tank = \(\frac { 32\pi }{ 3 } \)
⇒ \(\frac { 4 }{ 3 } { \pi r }^{ 3 }=\frac { 32\pi }{ 3 } \)
⇒ 4r³ = 32
⇒  r³ = \(\frac { 32 }{ 4 } \) = 8
⇒ r³ = 2³
⇒  r = 2
∴ Radius of the spherical tank is 2 unit

\(\left( \left[ (256)^{ \frac { -1 }{ 2 } } \right] ^{ \frac { -1 }{ 4 } } \right) ^{ 3 }\)  =  \((256)^{ \frac { -1 }{ 2 } \times \frac { -1 }{ 4 } \times 3 }\)   \([\because \frac { { a }^{ m } }{ { a }^{ n } } ={ a }^{ m-n }]\)
=  \((256)^{ \frac { 3 }{ 8 } }=({ 2 }^{ 8 })^{ \frac { 3 }{ 8 } }={ 2 }^{ 8\times \frac { 3 }{ 8 } }={ 2 }^{ 3 }=8\)

\(\left( 5^{ 3 } \right) ^{ \frac { 2 }{ 3 } }={ 5 }^{ 3\times \frac { 2 }{ 3 } }\) 
= 5² = 25  \(\left[ \because ({ a }^{ m })^{ n }={ a }^{ mn } \right] \)

Required polynomial equation is
\(\mathrm{p}(x)=a x^{2}+b x+c\)
Given one root is \(1+\sqrt{5}\)
another root is \(1-\sqrt{5}\)
Required equation is s (x- a)(x- b)
\( =(x-(1+\sqrt{5}))(x-(1-\sqrt{5})) \)
\( =(x-1-\sqrt{5})(x-1+\sqrt{5}) \)
\( =(x-1)^{2}-(\sqrt{5})^{2} \)
\( =x^{2}-2 x+1-5 \)
\( =x^{2}-2 x-4\)

Given \(\frac { { 3 }^{ 2n }{ 9 }^{ 2 }{ 3 }^{ -n } }{ { 3 }^{ 3n } } \) = 27
⇒  \(\frac { { 3 }^{ 2n-n }.{ 9 }^{ 2 } }{ { 3 }^{ 3n } }\)   = 27  [∵ a^m.aⁿ = a^m+n]
⇒  \(\frac { { 3 }^{ n }.({ 3 }^{ 2 })^{ 2 } }{ { 3 }^{ 3n } } \) = 27
⇒  3^n-3n (3⁴) = 27 \(\left[ \because \frac { a^{ m } }{ a^{ n } } ={ a }^{ m-n }\& ({ a }^{ m })^{ n }={ a }^{ mn } \right] \)
⇒  3^-2n.3⁴ = 27
⇒  3^-2n+4 = 3³
Equating the powers both sides we get
 -2n+4 = 3
⇒ -2n = 3-4 = -1
⇒ 2n = 1
⇒ n = \(\frac { 1 }{ 2 } \)